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Consider two closed (compact if needed) convex sets $E$ and $F$. Define the $\epsilon$ neighborhood of set $E$ as $$E_\epsilon = \cup_{x \in E} B_\epsilon (x),$$ where $B_\epsilon(x)$ is the open $\epsilon$-ball around $x$.

Now, given a $\delta>0$, is there always an $\epsilon>0$, such that $$E_\epsilon \cap F \subseteq (E\cap F)_\delta.$$


One counter example for $E$ not closed: enter image description here The intersection $E\cap F$ is the top right corner, and its $\delta$-neighborhood is just the $\delta$-ball. However, for any $\epsilon>0$, $E_\epsilon \cap F = F$, which would easily be out of $(E\cap F)_\delta$.


Some backgrounds: I started with the problem without the closedness and convexity assumption, and intuitively it makes sense. Then I started to see counter examples. With (almost) the strongest assumption, is the above statement correct?

One more assumption I can think of is to assume that $\mathrm{int}(E) \cap \mathrm{int}(F) \ne \emptyset$.

Thanks for any help / suggestions / hints!

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  • $\begingroup$ You are assuming that $E\cap F\neq \emptyset$, right? Otherwise, its $\delta$-neighbourhood would not be defined. $\endgroup$ Aug 15, 2022 at 18:23
  • $\begingroup$ Also, is this question in $\mathbb{R}^d$? Or in a more general vector space? $\endgroup$ Aug 15, 2022 at 18:26
  • $\begingroup$ Hey Keen, yes, I am assuming $E \cap F$ is non-empty. And yes, you can assume $\mathbb{R}^d$. Thanks for checking my question. $\endgroup$ Aug 15, 2022 at 20:40

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Assuming $X=\mathbb{R}^d$, I think the answer is true. Assume towards contradiction that the statement is not true. Then for every $n$ there exists an $x_n\in F$ such that $$d(x_n,E)<\frac{1}{n} \quad \text{and} \quad d(x_n,E\cap F)\geq \delta.$$

If $F$ is compact there exists an $x_\infty \in F$ such that $x_n\to x_\infty$. Since $E$ is closed and $d(x_n,E)\to 0$, we have $x_\infty \in E$. Hence, $x_\infty \in E\cap F$ while $d(x_\infty,E\cap F)\geq \delta$. This is clearly a contradiction.

My proof is weird because it just needs for the intersection not to be empty and the sets to be closed. Nowhere do I rely on convexity.

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  • $\begingroup$ Hey Keen, thanks a lot for answering the question. The only spot I would make change to your post is $x_{n_k} \to x_\infty$, we probably only have subsequence convergence for compact set. Other than this, everything looks right. Thank you so much. $\endgroup$ Aug 15, 2022 at 21:28
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If we allow $E$ and $F$ to be compact, and $E \cap F \neq \emptyset$ then the result is quite trivial in any metric space. In fact we don't need compactness - if one of them is bounded statement holds.

If $E$ was bounded, so would be $E_\epsilon$, so $E_\epsilon \cap F$ is bounded either way (as a subset of bounded set). Choose ball $B(x, r) \supseteq E_\epsilon \cap F$ and any point $p \in E\cap F$. For $\delta = r + d(x, p)$ we have $B(x, r) \subseteq B(p, \delta) \subseteq (F\cap E)_\delta$.

So sets don't even need to be closed.

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  • $\begingroup$ Hey Esgeriath, thanks a lot for trying to answer my question. I think maybe I did not pose the question well enough. I was asking given an arbitrary $\delta$ if there exists an $\epsilon$ that will satisfy $E_\epsilon \cap F \subseteq (E\cap F)_\delta$. I added a counter-example in the original post, and I hope it would clarify a little bit. Again thanks for the help! $\endgroup$ Aug 15, 2022 at 21:19
  • $\begingroup$ @Lyapunov1729 so you're asking about what conditions would need to hold in order to ensure your statement? I mean, you just gave counterexample yourself $\endgroup$
    – Esgeriath
    Aug 15, 2022 at 21:45
  • $\begingroup$ In the counter example, the set $E$ is not closed. That's why I added closed / compact assumption to the problem statement. $\endgroup$ Aug 15, 2022 at 21:52

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